A proton that has a mass m and is moving at 250 m/s in the + direction undergoes a head-on elastic collision with a stationary oxygen nucleus of mass 16m. Find the velocities of the proton and the oxygen nucleus after the collision

i know i have to consider that both momentum and KE are conserved but i don't know what to do (how to do it)

u1 = (m1-m2) •v/(m1+m2) – (m-16m) •250/(m+16m) = - 220.6 m/s,

Minus indicates that the proton is returned back due to the collision.
m1•v = m1•u1+m2•u2
u2 = (v+u1)m1/ m2 = 250 –(-220.6) •m/16•m = 29.4 m/s
in the direction of the initial direction of the proton.

To solve this problem, you will need to apply the principles of conservation of momentum and kinetic energy. Here's a step-by-step guide on how to approach the problem:

1. Write down the given information:
- Mass of the proton: m
- Initial velocity of the proton: 250 m/s in the + direction
- Mass of the oxygen nucleus: 16m
- Initial velocity of the oxygen nucleus: 0 m/s (stationary)

2. Apply the conservation of momentum:
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision.
Momentum (p) is calculated as the product of mass (m) and velocity (v).

Before the collision:
Momentum of the proton = m * 250 (in the + direction)
Momentum of the oxygen nucleus = 16m * 0 (stationary)
Total momentum before the collision = m * 250

After the collision:
Let the final velocity of the proton be v1 in the + direction
Let the final velocity of the oxygen nucleus be v2 in the opposite (-) direction
Momentum of the proton = m * v1 (in the + direction)
Momentum of the oxygen nucleus = 16m * v2 (in the - direction)
Total momentum after the collision = m * v1 - 16m * v2

Equate the total momentum before and after the collision:
m * 250 = m * v1 - 16m * v2

3. Apply the conservation of kinetic energy:
In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

Before the collision:
Kinetic energy (KE) is given by (1/2) * mass * velocity^2.
KE of the proton = (1/2) * m * (250)^2
KE of the oxygen nucleus = (1/2) * 16m * (0)^2 (stationary)
Total KE before the collision = (1/2) * m * (250)^2

After the collision:
KE of the proton = (1/2) * m * (v1)^2
KE of the oxygen nucleus = (1/2) * 16m * (v2)^2
Total KE after the collision = (1/2) * m * (v1)^2 + (1/2) * 16m * (v2)^2

Equate the total kinetic energy before and after the collision:
(1/2) * m * (250)^2 = (1/2) * m * (v1)^2 + (1/2) * 16m * (v2)^2

4. Solve the equations simultaneously to determine v1 and v2:
Solve the momentum equation (from step 2) for v2 in terms of v1 and substitute it into the kinetic energy equation (from step 3).
Then solve the resulting equation for v1.

Substitute the value of v2 from the momentum equation:
m * 250 = m * v1 - 16m * (250 - v1)

Simplify the equation:
250 = v1 - 16 * (250 - v1)

Expand and solve for v1:
250 = v1 - 16 * 250 + 16 * v1
250 = 17v1 - 4000
17v1 = 4250
v1 = 250

Substitute the value of v1 into the momentum equation to find v2:
m * 250 = m * 250 - 16m * v2
0 = - 16m * v2
v2 = 0

5. Calculate the final velocities:
From the analysis, we find that the final velocity of the proton (v1) is 250 m/s, and the final velocity of the oxygen nucleus (v2) is 0 m/s.

Therefore, after the collision, the proton continues to move in the + direction with a velocity of 250 m/s, and the oxygen nucleus remains stationary.